Question

If a particle moves with velocity u1 for some time and with velocity u2
for double of initial time along a
straight line, find the average speed for
the entire journey.
​

Answer 1

The average speed of the entire journey will be $\frac{u_{1}+2u_{2} }{3}$.

Explanation:

Required quantity: The average speed of the entire journey.

Formula Used:

Average Speed =$\frac{Total Distance Covered}{Total Time Taken}$

Distance= Speed X Time

Given:

The velocity of the particle in the first period of time: $u_{1}$

The velocity of the particle in the next period of time: $u_{2}$

The second interval of time= 2 X the first interval of time

Let the first interval of time for the velocity $u_{1}$ be: t

∴The next interval of time for the velocity $u_{2}$= 2t

Now to find the average speed we need to know the total distance. So we first try to find the distance covered by the particle in with velocity $u_{1}$.

Let the distance covered with velocity $u_{1}$ be s and with $u_{2}$ be s$_{2}$

we know, Distance= Speed X Time

therefore, s= $u_{1}$ X t

or s= $u_{1}t$...…..(1)

Also, for the distance s$_2$ we have velocity $u_{2}$ and time 2t.

similarly, s$_{2}$= $2u_{2}t$

So, now we have both the distances from which we can get the total distance and we have total time. Using these we can find the average velocity.

Average velocity= $\frac{Total Distance Covered}{Total Time Taken}$

putting the values we get,

Average velocity=$\frac{u_{1}t +2u_{2}t }{t+2t}$

Adding the value of time in the denominator we get,

Average velocity=  $\frac{u_{1}t +2u_{2}t }{3t}$

As we can see that there is nothing else to add so now we take t common from the numerator,

Average velocity= $\frac{(u_{1}+2u_2)t}{3t}$

now, canceling out t from the denominator and  numerator we get,

∴ Average velocity = $\frac{u_{1}+2u_{2} }{3}$